Cho $\Large I=\int\limits_{0}^{1}x\mathrm{ln}(2+x^2)\mathrm{d}x=a+b\ma

Cho $\Large I=\int\limits_{0}^{1}x\mathrm{ln}(2+x^2)\mathrm{d}x=a+b\mathrm{ln}2+c\mathrm{ln}3$ với $\Large a, b, c \in \mathbb{Q}.$ Tính giá trị của biểu thức $\Large T=a+b+c.$

• A.

  $\Large T=\dfrac{3}{2}.$

• B.

  $\Large T=\dfrac{7}{2}.$

• C.

  $\Large T=0.$

• D.

  $\Large T=2.$

Đặt $\Large \left\{\begin{align} & u=\mathrm{ln}(2+x^2)\\ & \mathrm{d}v=x\mathrm{d}x \end{align}\right.$ $\Large \Rightarrow \left\{\begin{align} &\mathrm{d}u=\dfrac{2x\mathrm{d}x}{2+x^2}\\ & v=\dfrac{x^2}{2} \end{align}\right.$ $\Large \Rightarrow I=\dfrac{x^2}{2}\mathrm{ln}(2+x^2)\bigg|_{0}^{1}-\int\limits_{0}^{1}\dfrac{x^3}{2+x^2}\mathrm{d}x=\dfrac{\mathrm{ln}3}{2}-A \ (1).$
Tính $\Large A=\int\limits_{0}^{1}\dfrac{x^3}{2+x^2}\mathrm{d}x.$
Đặt $\Large t=2+x^2 \Rightarrow \mathrm{d}t=2x\mathrm{d}x$ và $\Large x:0 \rightarrow 1$ thì $\Large  t:2\rightarrow 3.$
Suy ra: $\Large A=\int\limits_{0}^{1}\dfrac{x^2}{2+x^2}.x\mathrm{d}x=\int\limits_{2}^{3}\dfrac{t-2}{t}.\dfrac{\mathrm{d}t}{2}=\dfrac{1}{2}\int\limits_{2}^{3}\left(1-\dfrac{2}{t}\right)\mathrm{d}t=\dfrac{1}{2}(t-2\mathrm{ln}|t|)\bigg|_{2}^{3}=\dfrac{1}{2}-\mathrm{ln}\dfrac{3}{2}\ (2).$
Thay (2) vào (1) ta được: $\Large I=\dfrac{\mathrm{ln}3}{2}-\left(\dfrac{1}{2}-\mathrm{ln}\dfrac{3}{2}\right)=-\dfrac{1}{2}-\mathrm{ln}2+\dfrac{3}{2}\mathrm{ln}3$
Suy ra: $\Large a=-\dfrac{1}{2}; b=-1; c=\dfrac{3}{2} \Rightarrow T=0.$
$\Large \rightarrow$ đáp án C.